Find the slope of normal to the curve x = 1 – $a sin\theta$, y = $b cos^2\theta$ at $\theta$ = $\pi\over 2$.

Solution :

We have, x = 1 – $a sin\theta$, y = $b cos^2\theta$

$\implies$  $dx\over d\theta$ = $-a cos\theta$  and $dy\over d\theta$ = $-2b cos\theta sin\theta$

$\therefore$  $dy\over dx$ = $dy/d\theta\over dx/d\theta$ = $2b\over a$ $sin\theta$

$\implies$  $dy\over dx$  at $\pi\over 2$ = $2b\over a$

Hence, Slope of normal at $\theta$ = $\pi\over 2$ = $-a\over 2b$

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